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                        8.4 SORENSON’S METHOD
                                                                                    ζ
                                                          ζ 2                      May 10, 2005  16:28 241
                                                                                    2max
                                                    ∆S max
                          X 2

                                     X 1


                                                        θ    ∆S o
                                              i − 1  i        i + 1
                                                                                     ζ 2 = 0
                                                                                     ζ 1
                        Figure 8.5. Grid line construction – Sorenson’s method.


                        boundary, than the problem becomes overspecified or ill-posed. Therefore, we can
                        specify either the value of ∂x 1 /∂ξ 2 or of x 1 . However, if only one of these two
                        boundary conditions is specified then the converged solutions to Equations 8.26
                        and 8.27 often demonstrate grid-node clustering in some portions of the domain
                        and highly sparse node distributions in other regions.
                           Ideally, one would like to have complete freedom to choose x 1 and x 2 locations
                        on the boundaries and yet achieve orthogonal intersection (or at any other desired
                        angle) of the grid lines with the boundaries. The method of Sorenson [71] allows
                        precisely this freedom. The method is described in the next section.




                        8.4 Sorenson’s Method

                        8.4.1 Main Specifications

                        Sorenson’s method permits coordinate and coordinate-gradient specification for the
                        same variable x 1 or x 2 at two of the four boundaries of the domain. Thus, let ξ 2 = 0
                        (south) and ξ 2 = ξ 2max (north) be these two boundaries as shown in Figure 8.5. We
                        now define

                                                            2      2    0.5
                                                  s 0 =   x +   x                          (8.29)
                                                            1     2 ξ 2 =0
                        or, in the limit,

                                                                        0.5

                                                           2         2

                                             ds 0     ∂x 1      ∂x 2
                                                 =           +              .              (8.30)
                                            d ξ 2     ∂ξ 2       ∂ξ 2
                                                                        ξ 2 =0
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